Reservoir Engineering – Numericals

1 ) The production data and PVT given in Table 1 are relative to a reservoir of oil to undersaturated pore volume
constant. The production of water is negligible .

Also known are the following information :

• Compressibility water , cw = 3.0 10-6 psi - 1

• Compressibility training cf = 3.5 10-6 psi - 1
• Saturation in water irreducible Swi = 0.23

Tabella 1

Production time Pressure Oil cumulative production Gas cumulative production Rs Bo co

(month) (psia) (bblST) (MMft3SC) (ft3SC/bblST) (bbl/bblST) (psia-1)
0 7394 0 0 1550 1.866
6 6844 7961400 11539.8 1550 1.883 1.6564E-05
12 6044 22240080 32350.68 1550 1.919 2.1039E-05
18 5194 35870760 52185.24 1550 1.953 2.1193E-05

It requires you to :
a. Estimate the Oil Originally In Place
b . Calculate the recovery after 18 months of production .

2 ) If with a flow rate Qg = 40000 m3 / d after 8 hours it induces a disturbance pressure that extends for 132 m ,
which distance will be reached by the disturbance of pressure in a time equal to 12 hours in the event of delivering a
flow rate equal to Qg = 120,000 m3 / d :

 528 m
 396 m
 132 m
 264 m

3 ) A test in a gas well has provided the following results . Calculate the flow equation of the well and the AOF .
4 ) An interference test has been performed on an oil reservoir in order to confirm areal hydraulic
communication between two wells (active and observation). Provide an evaluation of the intra-well
permeability by using the following data and map.

3
Oil rate 250 m /day
Net pay 20 m
3 3
Oil formation volume factor 1.4 m r/m ST

Oil viscosity 1.2 cP

-4
Oil compressibility 3.2 10 1/bar
-5
Water compressibility 4.1 10 1/bar
-5
Formation compressibility 4.0 10 1/bar
Water saturation 0.20 (-)
Porosity 0.19 (-)
Lag time 26 hrs

Active Well

Observation Well

100 m
SCALE

5 ) A limit test is planned to investigate the reservoir area. Estimate the expected duration of the test.

1 km

k = 110 mD

 = 0.21
-4 -1
ct = 3 10 bar

 = 0.6 cP
rw = 6 in
9 ) If with a flow rate Qg = 40000 m3 / d after 8 hours it induces a disturbance pressure that extends for
163 m , which distance will be reached by the disturbance of pressure in a time equal to 12 hours in the
event of delivering a flow rate equal to Qg = 80000 m3 / d :

 326 m

 240 m

 163 m

 200 m

10 ) Based on the following well testing data on oil wells, indicate which well has the higher productivity. Data: Q 1
3 3
= 450 m /d; Q2 = 100 m /d; k1 = 100 mD; k2 = 50 mD

Well 1 Well 2

Elapsed time Pressure Elapsed time Pressure

hr bara hr bara
0.0 344.7 0.0 344.7
0.5 223.0 0.5 340.1
1.0 204.9 1.0 338.3
1.5 198.7 1.5 335.1
2.0 195.1 2.0 332.1
2.5 192.6 2.5 331.2
3.0 190.6 3.0 329.5
3.5 189.0 3.5 327.5
4.0 187.6 4.0 324.8
4.5 186.4 4.5 322.5
5.0 185.2 5.0 320.2
5.7 184.1 5.7 318.5
6.4 182.9 6.4 315.2
7.1 181.8 7.1 312.9
8.0 180.7 8.0 310.7
8.5 305.3
9.0 303.1
10.0 299.8

11 ) A gas field has an initial pressure of 300 bar and a volume of gas originally in place of 10 x 109 m3sc .
Assuming that the gas is ideal , estimate the amount of recoverable gas at a pressure drop of 40 bar . Considering the
fact that in fact it is a real gas , which correction should be made to the previous calculation ?

12 ) Looking at the curve of decline of an oil well in the chart, assessing which of the following coefficients of
decline is the correct one:
 3 % / month
 3 % / day
 5 % / month
 5 % / day
 10 % / month
 10 % / day

13 ) Consider a gas field dry . They are known, the following data :

GOOIP 2 Tscf

THP abandonment 150 psi

Wellbore average gas density at abandonment conditions 0.07 psi/ft

Reservoir depth 6500 feet

Initial reservoir pressure 2700 psi

Assuming the following values of Z :

Pressure Z Factor P/Z

(psia)

-------- -------

150 0.98764 151.8772

500 0.96065 520.48092

600 0.95351 629.25402

700 0.94665 739.44964

1000 0.92801 1077.5746

 1500 0.90451 1658.3565

2000 0.89201 2242.1273

2500 0.89091 2806.1196

2700 0.89347 3021.9258

3000 0.90018 3332.6668

Evaluating the final recovery factor and reserves associated with using the method of P / Z , please give the correct
answer of the following :

 RF = 50 % , Reserves = 1.2 TSCF

 RF = 71 % , Reserves = 1:42 TSCF
 RF = 73 % , Reserves = 1.46 TSCF
 RF = 79 % , Reserves = 1:58 TSCF
 RF = 95 % , Reserves = 1.9 TSCF

14 ) The figure below shows a study of the sensitivity of the well performance of a well taking oil obtained : Skin =
0 , 6 = Skin , Tubing OD = 4 1/2 " OD Tubing = 5 " . The working point to Skin = 0 and OD = 4 1/2 " is:

 5800 STB / D , 1900 psia

 4,000 STB / D , 2100 psia
 4400 STB / D , 2200 psia

15 ) Calculate the AOF of a gas well , based on the following input data : P * = 2400 psia , Pdew point = 2,400 psia .
BHP = 1200 psia flowing , Gas MMSCF rate = 20 / D . The equations of riferimeto are the following :
 Rawlins - Schellardt

Forcheimer approssimata

Vogel

16 ) In the process of the scale-up of the log curves , if within a cell of grid fall 60 % clay and 40% of sand
which have values of porosity average respectively equal to 0 for the clay and 26 % for sand , which will
have the petrophysical characteristics of the grid cell ?

a) NTG = 1 PHI = 17.3 %

b) NTG PHI = 0.4 = 13.5 %
c) NTG PHI = 0.6 = 26 %
d) NTG PHI = 0.4 = 10.4 %
e) NTG PHI = 0.4 = 26 %

17 ) Given that the well 1 is all oil bearing and the shaft 2 is all water bearing , where you can place the
oil water contact ( OWC ) ?

2502 m TVDSS

2557 m TVDSS

2568 m TVDSS

2585 m TVDSS
2595 m TVDSS
 a) Between 2557 m and 2568 m TVDSS TVDSS
b) Between 2557 m and 2595 m TVDSS TVDSS
c) Between 2502 m and 2557 m TVDSS TVDSS
d) Between 2557 m and 2585 m TVDSS TVDSS
e) Between 2568 m and 2585 m TVDSS TVDSS

18 ) Assuming that after 6 hours of production at a portatacostante equal to Qo = 400 m3 / d tildisturbo

pressure extends to 157 m from the well, which distance will be reached by the disturbance pressure
after 8 hours nell'oipotesi that the scope of the well equals Qo = 600 m3 / d?

a) 81 m
b) 157 m
c) 163 m
d) 181 m
e) 235 m

19 ) For an oil field that it was decided to produce with waterflooding was derived fractional flow curve
shown in Fig. Assume that the initial water saturation that is irreducible.

In the region of the reservoir where the oil is displaced by water saturation in water at the average
breakthrough will be approximately equal to:

a) 0.45
b) 0.55
c) 0.60
d) 0.70
20 ) We analyze the production data in the table by the material balance equation for the case of
undersaturated oil ( p > PSAT ). It assumes that the water production, the effect of the compressibility
of the rock and measurement errors are negligible. Evaluate the initial oil in place.

Time Tank Oil Solution Producing Cum Oil

Pressure FVF GOR GOR Produced

(date m/d/y) (psig) (RB/STB) (scf/STB) (scf/STB) (MMSTB)

1/1/2001 6000 1.55106 500 0

1/1/2002 4444 1.56088 500 500 2.51775

1/1/2003 3759 1.56782 500 500 4.27645

1/1/2004 3354 1.57327 500 500 5.64747

1/1/2005 2878 1.58166 500 500 7.73934

1/1/2006 2602 1.58795 500 500 9.29183

1/1/2007 2419 1.59293 500 500 10.5137

1/1/2008 2287 1.597 500 500 11.5076

1/1/2009 2194 1.59854 498 499 12.3363

1/1/2010 2160 1.58869 489 494 13.0812

1/1/2011 2131 1.58047 481 485 13.7891

1/1/2012 2106 1.5734 475 478 14.4653

a) 400 MMstb
b) 415 MMstb
c) 552 MMstb
d) 741 MMstb
e) 1018 MMstb

21 ) You want to estimate gas in place and the Recovery Factor final for a gas field by the method p / Z .
Are known, the volumes produced in the following table :
 Gp (MMscf) P/Z

100000 3299.34

500000 3166.033

1000000 2999.4

It is well known that the initial pressure of the reservoir is 3000 psi , the THP required minimum is 100
psi , the reservoir is 5700 feet deep and the average density of the gas in the well under the conditions
of THP limit is 0.158 psi / ft. Finally , use the table below to the properties of the gas :

Pressure Z Factor P/Z

(psia)

150 0.98764 151.8772

500 0.96065 520.4809

600 0.95351 629.254

700 0.94665 739.4496

1000 0.92801 1077.575

1500 0.90451 1658.357

2000 0.89201 2242.127

2500 0.89091 2806.12

2700 0.89347 3021.926

3000 0.90018 3332.667

As the source, including the following, the correct answer:

a) GOIP TSCF and RF = 8 = 75%

b) GOIP TSCF and RF = 10 = 68%
c) GOIP TSCF and RF = 12 = 66%
d) GOIP TSCF and RF = 14 = 71%
e) GOIP TSCF and RF = 6 = 79%

22 ) From a production test with constant flow we provide the following information:

compressibility total = 5 x 10-5 psi-1

oil flow = 400 bblST / day
OFVF = 1.26 BBLR / bblST
The pore volume of the associated field ( in m3 ) is closer to:

a) 5,200
b) 52,000
c) 520,000
d) 5,200,000

23 ) In a well just realized it was conducted a production test . The well was produced at a constant flow
rate of 300 bbl / d of oil . During dispensing , duration 3 days, and the next rise of the pressure (duration
48 hours) were harvested the following information :

Fluid properties and rock

Βo = 1.27 bblR/bblST h =37 ft

μo = 0.80 cP ct = 4.25 10-6 psi-1

 = 12%
(1 psi = 6895 Pa)
The radius of investigation (in meters ) reached during the build-up is in the range :

a) 140-160
b) 175-200
c) 225-255
d) 870-980

24 ) The figure below shows a study of the sensitivity of the well performance of a well taking oil
obtained : Skin = 0 , 6 = Skin , Tubing OD = 4 1/2 " OD Tubing = 5 " . The working point to Skin = 0 and OD
= 4 1/2 " is about ... ?

STB/D psia

Skin = 6 and 5” Tubing 5800 1900

Skin = 0 and 4 ½” Tubing 4400 2200

Skin = 0 and 5” Tubing 6200 2100

Skin = 6 and 4 1/2” Tubing 4000 2100

25 ) Assume an oil well has been produced at a constant rate Qo = 360 mST3/d and the following pressure
data have been recorded in time. Oil viscosity = 1.5 cP, net pay = 20 m, formation volume factor Bo =
1.26 bblr/bblst, total compressibility ct = 0.000145 bar-1, well radius rw = 0.108 m, porosity  = 0.3.

Calculate the reservoir permeability.

 Time Pressure
(min) (bar)
0 344.738
0.0015 344.719
0.004 344.685
0.012 344.58
0.0219 344.449
0.2409 341.746
0.6549 338.169
1.0797 335.904
2.403 332.486
5.9105 329.691
13.154 327.944
26.489 326.665
48.2661 325.644
107.4182 324.333
264.206 322.906
360 322.416

26 ) Given a gas reservoir with pi = 393.2 kg/cm² @ 3230 m ss, provide an interpretation of the
production data given below (Treservoir = 50°C; cf = 2 10-4 bar-1; Sw = 0.23).

P Gp Bg

[bar] [10³ m³] [m³/m³sc]

386,51 0 0,003030
380,41 34224 0,003076
360,02 1731188 0,003242
340,40 321560 0,003421
320,33 453407 0,003625
300,06 595970 0,003859
279,98 722142 0,004123
259,76 840803 0,004429
239,82 957118 0,004781
220,25 1084084 0,005186
199,75 1220589 0,005694
180,37 1325295 0,006277
159,86 1454676 0,007044
140,16 1599423 0,007988
120,34 1764047 0,00924
109,91 1853754 0,010077
27 ) Given a gas reservoir, provide an interpretation of the production data given below (Treservoir = 50°C)

P Bg Gp
[bar] [m³/m³sc] [10³ m³]
386,2 0,003032 0
359,7 0,003244 179412
320,1 0,003627 524692
279,8 0,004125 811140
239,7 0,004783 1225681
199,7 0,005695 1620668
159,9 0,007042 1928531

28 ) Production data in table 1 come from a saturated-oil reservoir with constant porous volume. Water
production is negligible.

PVT relevant data is also summarized in table 1, and Boi = 1,743 rb/stb.
Additionally, the following data is available:

 Water compressibility, cw = 3.0 10-6 psi-1

 Formation compressibility cf = 3.5 10-6 psi-1
 Irriducibile water saturation Swi = 0.23

TABLE 1

Time Pressure Rs Bo Bg Np Gp

(months) (psia) scf/stb rb/stb rb/scf stb MMscf

21 4500 1450 1,850 - 42686100 62102,52

24 4350 1323 1,775 0,000797 49004820 74224,17
30 4060 1143 1,670 0,000840 58079700 100758,69
36 3840 1037 1,611 0,000881 64420560 125109,45
42 3660 985 1,566 0,000916 69105420 145850,31
48 3480 882 1,523 0,000959 74046780 173060,37
54 3260 791 1,474 0,001015 77822280 198499,05
60 3100 734 1,440 0,001065 80657820 221018,31
66 2940 682 1,409 0,001121 83226060 242072,91
72 2900 637 1,382 0,001170 85713480 262195,47

Estimate the Original Oil In Place.